JEE Mains Model Paper 14

Let D = |

a	b	c
b	c	a
c	a	b

| = - (a3+b3+c3 - 3abc)
= - (a + b + c) (a2+b2+c2 - ab - bc - ca)
Now a2+b2+c2 - ab - bc - ca ≠ 0
∴ the given system of equations can have non-zero solution (x, y, z are not all zero)
If D = 0 , i.e. If a + b + c = 0
∴ Statement 1 is false
Again a2+b2+c2 - ab - bc - ca
=

1

2

[2a2+2b2+2c2 - 2ab - 2bc - 2ca]
=

1

2

[(a−b)2+(b−c)2 + (c−a)2] > 0
[∵ a, b, c are all different]
∴ a2+b2+c2 > ab + bc + ca
∴ Statement is true
∴ D holds